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IOSR Journal of Mathematics (IOSR-JM)
e-ISSN: 2278-5728, p-ISSN: 2319-765X. Volume 11, Issue 4 Ver. III (Jul - Aug. 2015), PP 65-69
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Jacobson Radical and On A Condition for Commutativity of
Rings
Dilruba Akter1, Sotrajit kumar Saha2
1
(Mathematics, International University of Business Agriculture and Technology, Bangladesh)
2
(Mathematics, Jahangirnagar University, Bangladesh)
Abstract: Some rings have properties that differ radically from usual number theoretic problems. This fact
forces to define what is called Radical of a ring. In Radical theory ideas of Homomorphism and the concept of
Semi-simple ring is required where Zorn’s Lemma and also ideas of axiom of choice is very important.
Jacobson radical of a ring R consists of those elements in R which annihilates all simple right R-module.
Radical properties based on the notion of nilpotence do not seem to yield fruitful results for rings without chain
condition. It was not until Perlis introduced the notion of quasi-regularity and Jacobson used it in 1945, that
significant chainless results were obtained.
Keywords: Commutativity, Ideal, Jacobson Radical, Simple ring, Quasi- regular.
I. Introduction
Firstly, we have described some relevant definitions and Jacobson Radical, Left and Right Jacobson
Radical, impact of ideas of Right quasi-regularity from Jacobson Radical etc have been explained with careful
attention. Again using the definitions of Right primitive or Left primitive ideals one can find the connection of
Jacobson Radical with these concepts. One important property of Jacobson Radical is that any ring 𝑅 can be
embedded in a ring 𝑆 with unity such that Jacobson Radical of both 𝑅 and 𝑆 are same. Another important result
is that any nontrivial ring 𝑅 is Jacobson semi-simple if and only if 𝑅 has been highlighted with proof.
Then we have discussed to condition for Commutativity of Rings. A well known theorem of Jacobson
asserts that if 𝑅 is a ring such that every element of 𝑅 is equal to some power of itself then 𝑅 is commutative. In
his proof he used axioms of choice. Herstein gave another proof of the same theorem without involving axiom
of choice. Considering the application of this theorem more elementary proof has been given.
1.1. Commutative ring.
Although ring addition is commutative, so that for every 𝑎, 𝑏 ∈ 𝑅, 𝑎 + 𝑏 = 𝑏 + 𝑎 ring multiplication
is not required to be commutative; 𝑎 − 𝑏 need not equal 𝑏 − 𝑎 for all 𝑎, 𝑏 ∈ 𝑅. Rings that also satisfy
commutativity for multiplication are called commutative ring.
Formally, let (𝑅, +,-) be a ring. Then (𝑅, +,-) is said to be a commutative ring if for every 𝑎, 𝑏 ∈ 𝑅, 𝑎 −
𝑏= 𝑏 − 𝑎 . That is, (𝑅, +,-) is required to be a commutative monoid under multiplication.
Example: The integers form a commutative ring under the natural operations of addition and multiplication.
1.2. Non commutative ring.
Non commutative ring is a ring whose multiplication is not commutative; that is 𝑎, 𝑏 ∈ 𝑅, 𝑎 − 𝑏 ≠
0 𝑎
𝑏 − 𝑎 Example: Let 𝑀 =
: 𝑎 and 𝑏 are real numbers .Then, (𝑀, +,-) is a non-commutative ring,
0 𝑏
without unity and without zero divisors.
1.3. Ring with unity.
A ring 𝑅 is said to be a ring with unity if it contains an element denoted by 1𝑅 or simply 1 such
that 𝑎. 1 = 1. 𝑎 ∀𝑎 ∈ 𝑅. The unity element 1 is called multiplicative identity.
Example: The set of all integers Z is a commutative ring with unity.
1.4. Division ring.
A ring 𝑅 in which the set 𝑅∗ of non-zero elements is a group with respect to the multiplication in 𝑅 is
called a division ring. Equivalently, 𝑅 is a division ring if every non-zero element of 𝑅 has a multiplicative
inverse in 𝑅.
Example: The rings 𝑄, 𝑅, 𝐶 are some examples of division rings.
DOI: 10.9790/5728-11436569
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Jacobson Radical and On A Condition for Commutativity of Rings
1.5. Nil and nilpotent.
An element 𝑥 is said to be nilpotent if there exists a positive integer n such that 𝑥 𝑛 = 0. A ring 𝑅 is
said to nil if every element 𝑥 ∈ 𝑅 is nilpotent, 𝑥 𝑛 = 0, where 𝑛 depends on the particular element 𝑥 of 𝑅.
II. Jacobson Radical
In Mathematics, more specifically ring theory, a branch of abstract algebra, the Jacobson radical of a
ring 𝑅 consists of those elements in 𝑅 which annihilate all simple right 𝑅 –module. The Jacobson radical of a
ring 𝑅 with 1 is defined as the radical ideal of 𝑅 with respect to the property that “A 2-sided ideal I is such that
1 − 𝑎 is a unit in 𝑅 for all 𝑎 ∈ 𝐼” and it is denoted by 𝐽(𝑅). In other words, 𝐽(𝑅) is the largest 2-sided ideal of 𝑅
such that 1 − 𝑎 is a unit for all 𝑎 € 𝐽(𝑅).
A computationally convenient notion when working with the Jacobson radical of a ring is the notion of
quasi-regularity. In particular, every element of ring’s Jacobson radical can be characterized as the unique right
ideal of a ring, maximal with respect to the property that each element is right quasi-regular.
The Jacobson radical of a ring is also useful in studying modulus over the ring. For instance, if 𝑈 is a
right 𝑅-module and 𝑉 is a maximal submodule of 𝑈. 𝐽(𝑅) is contained in 𝑉, where 𝑈. 𝐽(𝑅) denotes all products
of elements of 𝐽(𝑅) with elements in 𝑈. The Jacobson radical may also be defined for rings without unity.
2.1 Left Jacobson radical.
For any ring 𝑅 with 1, the intersection of all maximal left ideals of 𝑅 is called the left Jacobson radical
or simply the left radical of 𝑅 and is denoted by 𝐽𝑙 (𝑅) (In case 𝑅 is commutative, 𝐽𝑙 (𝑅) is the intersection of all
maximal ideals of 𝑅.)
Some examples, the left radical of a division ring is (0). More generally, the left radical of 𝑀𝑛 (𝐷) is
0 (∀𝑛 ∈ 𝑁) where 𝐷 is a division ring. The (left) radical of Z is (0). The (left) radical of a local ring is its
unique maximal ideal.
2.2. Right Jacobson radical.
For any ring 𝑅 with 1, the intersection of all maximal right ideals of 𝑅 is called the right Jacobson radical or
simply the right radical of 𝑅 and is denoted by 𝐽𝑟 (𝑅).
Note, (In case 𝑅 is commutative, 𝐽𝑟 (𝑅) is the intersection of all maximal ideal of 𝑅.)
Some examples, a) 𝐽𝑟 (𝑅) is a 2-sided ideal of R. b) 𝐽𝑟 𝑅 = {𝑥 ∈ 𝑅 │1 − 𝑥𝑦 a unit, ∀𝑦 ∈ 𝑅}
2.3. Maximal left ideal.
A left ideal I in 𝑅 is said to be a maximal left ideal in 𝑅 if 𝐼 ≠ 𝑅 and for a left ideal 𝐽 of 𝑅, 𝐼 ⊆ 𝐽 ⊆ 𝑅 ⇒ 𝐽 = 𝐼
or, 𝐽 = 𝑅, i.e., there are no left ideals strictly in between 𝐼 and 𝑅.
2.4. Minimal left ideal.
A left ideal 𝐼 in 𝑅 is said to be a minimal left ideal in R if 𝐼 ≠ 0 and for a left ideal 𝐼 of 𝑅, 0 ⊆ 𝐽 ⊆ 𝐼 ⇒ 𝐽 =
(0) or, 𝐽 = 𝐼 i.e., there are no left ideals strictly in between (0) and 𝐼.
Remark, Maximal (resp. minimal) right/2-sided ideals are defined in exactly the same way as above.
2.5. Right quasi-regular ring.
A ring 𝑅 is right quasi-regular if every element in it is right quasi-regular.
2.5.1. Right quasi-regular right ideal.
If 𝐼 is a right ideal of a ring 𝑅 and if every element of 𝐼 is right quasi-regular, then 𝐼 is a right quasi-regular right
ideal (or right quasi-regular left ideal or two sided ideal).
2.6. Right primitive.
𝑅 is right primitive if 𝑅 contains a maximal right ideal 𝑀 such that (𝑀: 𝑅) = 0
2.6.1. Right primitive ideal.
An ideal (two-sided) 𝑃 of 𝑅 is a right primitive ideal if 𝑅/𝑃 is right primitive.
III. Wedderburn’s theorem
If 𝐴 is a simple ring unit 1 and minimal left ideal 𝐼, then 𝐴 is isomorphic to the ring of 𝑛 × 𝑛 matrices
over a division ring.
Proof. Let 𝐷 be a division ring and 𝑀(𝑛, 𝐷) be the ring of matrices with entries in 𝐷. It is not hard to show that
every left ideal in 𝑀 𝑛, 𝐷 takes the form
DOI: 10.9790/5728-11436569
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Jacobson Radical and On A Condition for Commutativity of Rings
{𝑀 ∈ 𝑀(𝑛, 𝐷)│The n1,…. 𝑛𝑘 – th columns of 𝑀have zero entires}, for some fixed n1,…. 𝑛𝑘 ⊂ {1, … , 𝑛}. So a
minimal ideal in 𝑀(𝑛, 𝐷) is of the form {𝑀 ∈ 𝑀(𝑛, 𝐷)│All but the k − th columns have zero entries}, for a
given 𝑘. In other words, if 𝐼 = (𝑀 𝑛, 𝐷 ) 𝑒 where 𝑒 is the idempotent matrix with 1 in the(𝑘, 𝑘) entry and zero
elsewhere.
Also,𝐷
is isomorphic to 𝑒 𝑀 𝑛, 𝐷 𝑒. The left ideal 𝐼 is a minimal left ideal,
then = 𝑀 𝑛, 𝐷 𝑒 𝑒 𝑘, 𝑘 𝐷 𝑒 𝑀 𝑛, 𝐷 𝑒 . 𝐼 over 𝑒 𝑀 𝑛, 𝐷 𝑒 , and the ring 𝑀(𝑛, 𝐷) is clearly isomorphic
to the algebra of homomorphisms on this module. The above example suggests the following lemma.
3.1.Lemma.
A is a ring with identity 1 and an idempotent element 𝑒 where 𝐴𝑒𝐴 = 𝐴. Let 𝐼 be the left ideal 𝐴𝑒,
considered as a right module over 𝑒𝐴𝑒. Then 𝐴 is isomorphic to the algebra of homomorphisms on 𝐼, denoted
by 𝐻𝑜𝑚(𝐼).
Proof. We define the left regular representation 𝜙: 𝐴 → 𝐻𝑜𝑚 𝐼 by 𝜙 𝑎 𝑚 = 𝑎𝑚 for 𝑚 ∈ 𝐼. 𝜙 is injective
because if 𝑎. 𝐼 = 𝑎𝐴𝑒 = 0, then 𝑎𝐴 = 𝑎𝐴𝑒𝐴 = 0, which implies 𝑎 = 𝑎. 1 = 0. For surjectivity, let 𝑇 ∈ 𝐻𝑜𝑚(𝐼).
Since 𝐴𝑒𝐴 = 𝐴, the unit 1 can be expressed as 1 = 𝑎𝑖 𝑒𝑏𝑖 . So 𝑇 𝑚 = 𝑇 1. 𝑚 = 𝑇 𝑎𝑖 𝑒𝑏𝑖 𝑚 =
𝑇 𝑎𝑖 𝑒𝑒𝑏𝑖 𝑚 = 𝑇 𝑎𝑖 𝑒 𝑒𝑏𝑖 𝑚 = [ 𝑇 𝑎𝑖 𝑒 𝑒𝑏𝑖 ]𝑚. Since the expression[ 𝑇 𝑎𝑖 𝑒 𝑒𝑏𝑖 ] does not depend on 𝑚,
𝜙 is surjective. This proves the lemma. Wedderburn's theorem follows readily from the lemma3.1.
IV. Zorn’s Lemma
A partially ordered non-empty set in which every chain is bounded above (resp. below) has a maximal
(resp.minimal) element.
Proof. A non-empty set 𝑋 with a partial order ′ ≤ ′ is called a poset, i.e. ′ ≤ ′, satisfies the following .
Reflexive. 𝑥 ≤ 𝑥, ∀𝑥 ∈ 𝑋 ,
Anti-symmetry. 𝑥 ≤ 𝑦 and 𝑦 ≤ 𝑥 ⇒ 𝑥 = 𝑦,
Transitivity. 𝑥 ≤ 𝑦 and 𝑦 ≤ 𝑧 ⇒ 𝑥 ≤ 𝑧 .
A subject 𝑌 of 𝑋 is called a chain or totally ordered if any two elements of 𝑌 are comparable, i.e., given 𝑥, 𝑦 ∈ 𝑌
either 𝑥 ≤ 𝑦 and 𝑦 ≤ 𝑥 . In particular, given finitely many elements 𝑦1 , … … … … , 𝑦𝑛 in 𝑌, there is a
permutation 𝜎 of 1,2, … … 𝑛 such that 𝑦𝜎(1) ≤ … … . . ≤ 𝑦𝜎 𝑛 .
A subset 𝐴 of 𝑋 is said to be bounded above (resp. below) if there is an 𝛼 ∈ 𝑋 such that 𝑎 ∈ 𝛼 ; 𝛼 𝜖 𝑎 ∀𝑎 ∈
𝐴 . Such an 𝛼 is called an upper (resp. a lower) bound for 𝐴. It need not belong to 𝐴. A subset 𝐴 of 𝑋 is said to
have a maximal (resp. minimal) element if there is an 𝑎 ∈ 𝐴 such that ≰ 𝑥 ; 𝑥 ≰ 𝑎 ∀ 𝑎 ∈ 𝐴, 𝑥 ≠ 𝑎 . Note that
a maximal (resp. minimal) element need not exists or it need not be an upper (resp. lower) bound when exists or
it need not be unique.
5.1. Lemma. (Andrunakievic).
Let A be an ideal of a ring 𝑅, 𝐵 be an ideal of 𝐴. Also let 𝐵 ∗ be the ideal of 𝑅 generated by 𝐵. Then
∗3
𝐵 ⊆ 𝐵.
Proof. 𝐵 ∗3 ⊆ 𝐴𝐵 ∗ 𝐴 = 𝐴 𝐵 + 𝑅𝐵 + 𝐵𝑅 + 𝑅𝐵𝑅 𝐴 ⊆ 𝐴𝐵𝐴 ⊆ 𝐵
5.2. Lemma.
If 𝐽 is the Jacobson radical of a ring 𝑅 and 𝑇 is an ideal of 𝑅, then thinking of 𝑇 as a ring, the Jacobson radical
of 𝑇 = 𝐽 ∩ 𝑇 . In particular, if 𝑅 is semi-simple, then so is 𝑇.
Proof. Let 𝐴 be the Jacobson radical of 𝑇 and 𝐴∗ be the ideal of 𝑅 generated by 𝐴. By Lemma 5.1, 𝐴∗3 ⊆ 𝐴 .
Now 𝐴∗ is an ideal of 𝑅 and since it is in , it is right quasi-regular.
If 𝑅 is semi-simple, then 𝐴∗3 = 0, 𝐴∗ is a nilpotent ideal of 𝑅, 𝐴∗ = 0, therefore 𝐴 = 0, 𝑇 is semi-simple. If 𝑅
is not semi-simple, then (𝑇 + 𝐽)/𝐽 is an ideal of the semi simple ring 𝑅/𝐽. Thus, (𝑇 + 𝐽)/𝐽 is semi-simple.
𝑇+𝐽
But 𝐽 ≅ 𝑇 𝑇 ∩ 𝐽 . Thus (𝑇 + 𝐽)/𝐽 is semi-simple. Now, 𝑇 ∩ 𝐽 is an ideal of 𝑇 and it is right quasi-regular.
Therefore, 𝑇 ∩ 𝐽 ⊆ 𝐴. If ∩ 𝐽 ≠ 𝐴 , then 𝐴/(𝑇 ∩ 𝐽) is a non-zero right quasi-regular ideal of 𝑇/(𝑇 ∩ 𝐽) . This is
impossible. Therefore, 𝑇 ∩ 𝐽 = 𝐴.
Remark. Lemma5.2 tells us that every ideal of a Jacobson radical ring is itself Jacobson Radical.
5.3. Lemma.
Any ring 𝑅 can be embedded in a ring 𝑆 with unity such that, Jacobson radical of 𝑅 = Jacobson radical of 𝑆.
Proof. If 𝑅 has a unity, then we merely take 𝑆 = 𝑅. If 𝑅 does not have a unity, then we adjoin one in
the standard way by taking ordered pairs of elements of 𝑅 and integers. Call this large ring 𝑆. Then 𝑅 is an
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Jacobson Radical and On A Condition for Commutativity of Rings
ideal of 𝑆 and 𝑆/𝑅 ≅ integers. Since 𝑆/𝑅 is semi-simple, 𝑅 must contains 𝐽𝑆 , the Jacobson radical of 𝑆 , on the
other hand, by lemma 5.2, 𝐽𝑅 the Jacobson radical of 𝑅, is equal to 𝐽𝑆 ∩ 𝑅. Since 𝐽𝑆 ⊆ 𝑅, we have 𝐽𝑆 = 𝐽𝑅 .
5.4. Lemma.
A simple non-trivial ring 𝑅 is Jacobson semi-simple if and only if 𝑅 has some maximal right ideals.
Proof. If 𝑅 is simple, non-trivial (i.e.𝑅2 ≠ 0), and not Jacobson radical, then there exists an element 𝑥 in 𝑅
which is not right quasi-regular. Then, 𝑅 contains maximal right ideals.
Conversely, if 𝑅 is simple, non trivial, and contains a maximal right ideal 𝑀, take 𝑥 in 𝑅, 𝑥 not in 𝑀. Consider
𝑥𝑅 Since 𝑀 is maximal, 𝑥𝑅 cannot be in 𝑀. To see this, consider {𝑦: 𝑦𝑅 ⊆ 𝑀} .
This set is clearly a right ideal of 𝑅, and it certainly contains 𝑀. However, it cannot be all of 𝑅 because 𝑅𝑅 is a
non-zero ideal of 𝑅 and must be equal to because 𝑅 is simple. Thus 𝑅𝑅 ⊈ 𝑀. Consequently, {𝑦: 𝑦𝑅 ⊆ 𝑀} .
must be equal to 𝑀. Then if ⊆ 𝑀 , 𝑥 is in 𝑀.
Since we selected 𝑥 not in 𝑀, 𝑥𝑅 is not in 𝑀. Then 𝑅 equals the right ideal generated by 𝑀 and 𝑥𝑅. In
particular, there must exist 𝑚 in 𝑀 and 𝑥 ′ in 𝑅, such that 𝑥 = 𝑚 + 𝑥𝑥 ′ . Let 𝑅𝑥 = ({𝑎: 𝑥𝑎𝜖𝑀} . This is a right
ideal. We observe that {𝑐 − 𝑥 ′ 𝑐} is contained in 𝑅𝑥 for 𝑥 𝑐 − 𝑥 ′ 𝑐 = 𝑥 − 𝑥𝑥 ′ = 𝑚𝑐 is in 𝑀. Now if 𝑅 is
Jacobson radical, then −𝑥 ′ is right quasi-regular and there exist an element 𝑧. Such that −𝑥 ′ + 𝑧 − 𝑥 ′ 𝑧 = 0 .
Then 𝑥 ′ = 𝑧 − 𝑥 ′ 𝑧 = 0 and this in 𝑅𝑥 . Therefore, 𝑥 ′ 𝑐 ∈ 𝑅𝑥 for every 𝑐. Then 𝑐 is in 𝑅𝑥 for every 𝑐 ,𝑅𝑥 = 𝑅.
However, in that case 𝑥𝑅 ⊆ 𝑀, a contradiction. Therefore, −𝑥 ′ cannot be right quasi-regular, 𝑅 is not Jacobson
radical and therefore, 𝑅 is Jacobson semi-simple.
Remark. If a simple ring is trivial, then it is nilpotent and therefore Jacobson Radical.
V. On a condition for Commutivity of Rings.
6.1. Jacobson Theorem.
Let 𝑅 be a ring such that every element in 𝑅 is a power of itself. Then 𝑅 is commutative.
Proof.
To proves this theorem, we first prove the three important lemmas.
6.1.1. Lemma.
Let 𝑅 be a non-commutative ring such that 𝑥 ∈ 𝑅, there exists a positive integer 𝑛(𝑥) > 1, such that
𝑥 𝑛 (𝑥) = 𝑥 . Then 𝑅 contains a non-commutative subring of prime characteristic.
Proof. By non-commutative, there exists an 𝑥 in 𝑅, such that 𝑥 is not in the centre of 𝑅 , so it will sufficient to
show that 𝑥, 𝑦 ∈ 𝑅 such that 𝑝𝑥 = 𝑝𝑦 = 0, where 𝑝 is a prime, and 𝑥𝑦 ≠ 𝑦𝑥 . Suppose that 𝑥 𝑛 = 𝑥, (2𝑥)𝑚 =
2𝑥,
then 𝑆 = 𝑛 − 1 𝑚 − 1 + 1, we have (2𝑥)𝑠 = 2𝑠 𝑥 𝑠 = 2𝑠 𝑥 = 2𝑥, whence 2𝑠 − 2 𝑥 = 0.
Suppose that 𝑡𝑥 = 0 with │𝑡│ minimal, then since 𝑅 contains no nilpotent, 𝑡 is square-free; whence 𝑥 is the
sum of elements each of prime characteristic. Clearly, 𝑝𝑥 = 𝑝𝑦 = 0, with 𝑝 and 𝑞 coprime implies that 𝑥𝑦 =
𝑦𝑥 = 0; whence the result follows.
6.1.2. Lemma.
Let 𝑅 be a ring of prime characteristic satisfying the conditions of lemma 6.1.1. Then 𝑅 contains a finite
non-commutative subring.
Proof. Each 𝑥 belonging to 𝑅 generates a subring which is the direct product of finite fields. Take some finite
field 𝐹, of order 𝑃 𝑆 , generated by an 𝑥 which is not in the centre of 𝑅 . If 𝑅 is non-commutative and 𝑥 is not in
the centre of 𝑅 then 𝑥 ∈ 𝑥 𝑛 𝑥 −1 𝑅 𝑥 𝑛 𝑥 −1
and, since
𝑅 = 𝑥 𝑛 𝑥 −1 𝑅 𝑥 𝑛 𝑥 −1 ⊕ 𝑥 𝑛 𝑥 −1 − 1 𝑅(𝑥 𝑛 𝑥 −1 − 1) ,
𝑥 is not in the centre of 𝑥 𝑛 𝑥 −1 𝑅 𝑥 𝑛 𝑥 −1 ; whence, by considering 𝑥 𝑛 𝑥 −1 𝑅 𝑥 𝑛 𝑥 −1 , we may assume that the
identity for 𝐹 is the identity for 𝑅.
Letting 𝑦 be an element of 𝑅 which does not commute with 𝑥, we have
𝑠
𝑦 𝑇𝑥𝑃 − 𝑇𝑥 = 𝑦 𝑇𝑥 − 𝐼𝜆1 𝑇𝑥 − 𝐼𝜆2 … … 𝑇𝑥 − 𝐼𝜆𝑃 𝑠 = 0,
Where 𝜆1 , … … 𝜆𝑃 𝑠 , are the elements of 𝐹 , 𝑇𝑥 is the mapping 𝑅 → 𝑅 defined by 𝑟 𝑇𝑥 = 𝑟𝑥 − 𝑥𝑟 and 𝐼𝜆𝑖 with
𝜆𝑖 ∈ 𝐹 is defined by 𝑟 𝐼𝜆𝑖 = 𝜆𝑖 𝑟. With 𝜆1 = 0 , we have an 𝑖 such that
𝑧 = 𝑦 𝑇𝑥 − 𝜆1 … … … … (𝑇𝑥 − 𝐼𝜆𝑖−1 ) ≠ 0,
and
𝑧 𝑇𝑥 − 𝐼𝜆𝑖 = 0, 𝜆𝑖 ≠ 0;
Whence the subring generated by 𝑥 and 𝑧 is finite.
DOI: 10.9790/5728-11436569
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Jacobson Radical and On A Condition for Commutativity of Rings
6.1.3.Lemma.
Let 𝑅 be a finite non-commutative ring such that every element of 𝑅 is a power of itself such that every
element of 𝑅 is a power of itself. Then 𝑅 contains a non-commutative division ring.
Proof.
Assume that 𝑆 is a minimal non-commutative subring of 𝑅. Then 𝑅, itself considered as a ring, contains
no zero divisor; for if 𝑎, 𝑏 ∈ 𝑆 with 𝑎𝑏 = 0, 𝑎 ≠ 0, 𝑏 ≠ 0 .
We have, 𝑆 = 𝑎𝑛 𝑎 −1 𝑆𝑎𝑛 𝑎 −1 ⊕ 𝑎𝑛 𝑎 −1 − 1 𝑆 𝑎𝑛 𝑎 −1 − 1 ,
With 𝑎 ∈ 𝑎𝑛 𝑎 −1 𝑆𝑎𝑛 𝑎 −1 and 𝑏 ∈ 𝑎𝑛 𝑎 −1 − 1 𝑆 𝑎𝑛 𝑎 −1 − 1 , which contradicts the minimality of 𝑆, since
the commutativity of both 𝑎𝑛 𝑎 −1 𝑆𝑎𝑛 𝑎 −1 and 𝑎𝑛 𝑎 −1 − 1 𝑆 𝑎𝑛 𝑎 −1 − 1 implies that 𝑆 is commutative.
Hence lemma 6.1.1, 6.1.2, and 6.1.3 together with Wedderburn’s theorem prove the Jacobson theorem.
VI. Conclusion
Looking the problem in difficult way lemma6.1.1 depicts that if 𝑅 be a non-commutative ring such that for
every 𝑥 belonging in 𝑅 there exists a positive integer 𝑛(𝑥) ≥ 1 such that 𝑥 𝑛 (𝑥) = 𝑥 then 𝑅 contains a noncommutative subring of prime characteristic.
Lemma 6.1.2 ensures that if 𝑅 be a ring of prime characteristic satisfying the conditions of lemma 6.1.1, then
𝑅 contains a finite non-commutative subring.
Lemma 6.1.3 also ensures that if 𝑅 be a finite non-commutative ring such that every element of 𝑅 is a power of
itself then 𝑅 contains a non-commutative division ring .
Lemma 6.1.1, 6.1.2 and 6.1.3 together with Wedderburn’s theorem establishes that, if 𝑅 be a ring with each
element being power of itself then 𝑅 must be commutative.
References
[1].
[2].
[3].
[4].
C. Musili, Introduction to Rings and Modules, University of Hyderabad, Norsha publishing house, (1999).24-26.
D.W. Henderson, A short proof of Wedderburn's theorem, Amer. Math. Monthly 72 (1965). 385-386
N. J. Divinsky, Ring and Radical, University of British Columbia, University of Toronto press, (1965). 91-115.
J.W.Wamsley, On a condition for commutativity of rings, J. LONDON MATH Soc.(2), 4 (1971).331-332
DOI: 10.9790/5728-11436569
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